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authorOpheliar99 <>2010-07-04 04:57:02 +0000
committerbnewbold <bnewbold@adelie.robocracy.org>2010-07-04 04:57:02 +0000
commitd18882b76339230566361b2a402bc5f543da6675 (patch)
tree882944cc902bf2c4d7fc6b8a30a2f49c35c17aaf
parent27194cc662f73877a413955c4c5f01761e2c8b08 (diff)
downloadafterklein-wiki-d18882b76339230566361b2a402bc5f543da6675.tar.gz
afterklein-wiki-d18882b76339230566361b2a402bc5f543da6675.zip
posted solutions of 2 and 3 in pset2
-rw-r--r--Problem Set 2.page3
1 files changed, 2 insertions, 1 deletions
diff --git a/Problem Set 2.page b/Problem Set 2.page
index 1cbcce1..06528fc 100644
--- a/Problem Set 2.page
+++ b/Problem Set 2.page
@@ -85,7 +85,8 @@ $m = -2 : a_m = \frac{1}{4 \sqrt{2\pi}} \times 2\pi = \sqrt{2\pi}/4$,
Then,
-$\sum |a_n|^2 = {\sqrt{2\pi}/4}^2 + {- \sqrt{2\pi}/2}^2 + {\sqrt{2\pi}/4}^2 = \frac{3 \pi}{4}$
+$\sum |a_n|^2 = {\sqrt{2\pi}/4}^2 + {- \sqrt{2\pi}/2}^2 + {\sqrt{2\pi}/4}^2 = \frac{3 \pi}{4}$.
+
And, it was shown in Prob 2 that $\int_0^{2\pi} \sin^4(x) dx = \frac{3 \pi}{4}$.
Therefore,